FACTOR: One number is said to be a factor of another when it divides the other exactly. Thus, 6 and 7 are factors of 42.
Question - Find the greatest number which is that when 76, 151 and 226 are divided by it, the remainders are all alike. Find also the common remainder.
Solution : Let k be the remainder, then the number (76 – k), (151 – k) and (226-k)
are exactly divisible by the required number.
Now, we know that if two numbers are divisible by a certain number, then their difference . hence, the numbers (151 – k ) – (76 – k), (226 – k) – (151 – k) and (226 –k) – (76 – k) or 75, 75 and 150 are divisible by the required number.
Therefore, the required number = HCF of 75, 75 and 150 =75.
Question - The numbers 11284 and 7655, when divided by a certain number of three digits, leave the same remainder. Find that numbers of three digits.
Solution: The required number must be a factor of (1128 – 7655) or 3629.
Now, 3629 = 19 x 191
Therefore, 191 is the required numbers.
Question - The product of two numbers is 7168 and their HCF is 16; find the numbers.
Solution: The numbers must be multiples of their number HCF. So, let the numbers be 16a and 16b where a and b are two numbers prime to each other.
∴ 16 a x 16 b = 7168 or, ab = 28
Now, the pairs of numbers whose product is 28 are 28, 1; 14 , 2; 7, 4.
14 and 2 which are not prime to each other should be rejected.
Hence, the required numbers are 28 x 16, 1 x 16; 7 x 16 ; 7 x 16, 4 x 16 or, 448, 16 ; 112, 64.
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COMMON FACTOR: A common factor of two or more numbers is a number that divides each of them exactly. Thus, 3 is a common factor of 9, 18, 21, and 33.
HIGHEST COMMON FACTOR: HCF of two or more numbers is the greatest number that divides each of them exactly. Thus, 6 is the HCF of 18 and 24. Because there is no number greater than 6 that divides both 18 and 24.
To find the HCF of two or more numbers:
Rule 1
Break the given numbers into prime factors and then find the product of all the prime factors common to all the numbers. The product will be the required HCF.Example: Find the HCF of 42 and 70.
Solution: 42 = 2 × 3 × 7
70 = 2 × 5 × 7
HCF = 2 × 7 = 14
Example: Find the HCF of 1365, 1560 and 1775.
Example: Find the HCF of 1365, 1560 and 1775.
Solution: 1365 = 3 ×5 × 7 ×13
1560 = 2 x 2 x 2 x 3 x 5 x 13
1755 = 3x 3 x 3 x 5 x 13
HCF = 3 x 5x 13 = 195
Rule 2 (METHOD OF DIVISON)
Divide the greater number by the smaller number, divide the divisor by the remainder by the next remainder and so on until no remainder is left. The last divisor is the required HCF.NOTE: The above rule is for finding the HCF of numbers is based on the following two principles:
- Any number which divides a certain number also divides any multiple of that number ; for example 6 divides 18 therefore, 6 divides any multiple of 18 .
- Any number which divided each of the number also divides their sum , their difference and the sum and difference of any multiple of that numbers.
To find the HCF of more than two numbers:
Rule
Find the HCF of any of the two numbers and then find the HCF of this HCF and the third number and so on. The HCF will be the required HCF.Example : Find the HCF of 1365, 1560 and 1755.
HCF of vulgar fractions:
Definition: The HCF of two or more fractions is the highest fraction which exactly divisible by each of the fractions.
Rule: First express the given fractions in their lowest terms:
Note: the HCF of a number of fractions is always a fraction ( but this is not the true LCM).
MISCELLANEOUS EXAMPLES ON HCF
- What is the greatest number that will divide 2400 and 1810 and leave remainders 6 and 4 respectively?
- What is the greatest number will be divide 38, 45 and 52 and leave as remainder 2, 3 and respectively?
- Find the greatest number which will divide 410, 751 and 1030 so as to leave remainder 7 in each case?
So the required number is 31.
Question - Find the greatest number which is that when 76, 151 and 226 are divided by it, the remainders are all alike. Find also the common remainder.
Solution : Let k be the remainder, then the number (76 – k), (151 – k) and (226-k)
are exactly divisible by the required number.
Now, we know that if two numbers are divisible by a certain number, then their difference . hence, the numbers (151 – k ) – (76 – k), (226 – k) – (151 – k) and (226 –k) – (76 – k) or 75, 75 and 150 are divisible by the required number.
Therefore, the required number = HCF of 75, 75 and 150 =75.
Question - The numbers 11284 and 7655, when divided by a certain number of three digits, leave the same remainder. Find that numbers of three digits.
Solution: The required number must be a factor of (1128 – 7655) or 3629.
Now, 3629 = 19 x 191
Therefore, 191 is the required numbers.
Question - The product of two numbers is 7168 and their HCF is 16; find the numbers.
Solution: The numbers must be multiples of their number HCF. So, let the numbers be 16a and 16b where a and b are two numbers prime to each other.
∴ 16 a x 16 b = 7168 or, ab = 28
Now, the pairs of numbers whose product is 28 are 28, 1; 14 , 2; 7, 4.
14 and 2 which are not prime to each other should be rejected.
Hence, the required numbers are 28 x 16, 1 x 16; 7 x 16 ; 7 x 16, 4 x 16 or, 448, 16 ; 112, 64.
Read all Quantitative Aptitude Shortcuts
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